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How much gas is present could be specified by giving the mass instead of the chemical amount of gas. Therefore, an alternative form of the ideal gas law may be useful. The chemical amount, n (in moles), is equal to total mass of the gas (m) (in kilograms) divided by the molar mass, M (in kilograms per mole): =.
In chemistry, the molar mass (M) (sometimes called molecular weight or formula weight, but see related quantities for usage) of a chemical compound is defined as the ratio between the mass and the amount of substance (measured in moles) of any sample of the compound. [1] The molar mass is a bulk, not molecular, property of a substance.
This formula is stated as: =, where: Rate 1 is the rate of effusion for the first gas. (volume or number of moles per unit time). Rate 2 is the rate of effusion for the second gas. M 1 is the molar mass of gas 1 M 2 is the molar mass of gas 2.
The other equation of state of an ideal gas must express Joule's second law, that the internal energy of a fixed mass of ideal gas is a function only of its temperature, with = (,). For the present purposes it is convenient to postulate an exemplary version of this law by writing:
It is an equation of state that relates the pressure, temperature, and molar volume in a fluid. The equation modifies the ideal gas law in two ways: first, it considers particles to have a finite diameter (whereas an ideal gas consists of point particles); second, its particles interact with each other (unlike an ideal gas, whose particles move ...
In thermodynamics, the specific volume of a substance (symbol: ν, nu) is the quotient of the substance's volume (V) to its mass (m): = It is a mass-specific intrinsic property of the substance. It is the reciprocal of density ρ and it is also related to the molar volume and molar mass:
The ideal gas equation can be rearranged to give an expression for the molar volume of an ideal gas: = = Hence, for a given temperature and pressure, the molar volume is the same for all ideal gases and is based on the gas constant: R = 8.314 462 618 153 24 m 3 ⋅Pa⋅K −1 ⋅mol −1, or about 8.205 736 608 095 96 × 10 −5 m 3 ⋅atm⋅K ...
This is illustrated in the image here, where the balanced equation is: CH 4 (g) + 2 O 2 (g) → CO 2 (g) + 2 H 2 O (l) Here, one molecule of methane reacts with two molecules of oxygen gas to yield one molecule of carbon dioxide and two molecules of liquid water. This particular chemical equation is an example of complete combustion.